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Question Number 141776 by 7770 last updated on 23/May/21

The volume of a sphere is increasing at the constant rate of 10cm^3 /sec. Calculate the rate of increase of the surface area at the instant when the radius is 5cm.What is the radius of the sphere when the surface area is increasing at 2cm^2 /sec. Any step by step solution pls.

$$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{volume}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{sphere}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{increasing}} \\ $$$$\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{constant}}\:\boldsymbol{\mathrm{rate}}\:\boldsymbol{\mathrm{of}}\:\mathrm{10}\boldsymbol{\mathrm{cm}}^{\mathrm{3}} /\boldsymbol{\mathrm{sec}}. \\ $$$$\boldsymbol{\mathrm{Calculate}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{rate}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{increase}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\: \\ $$$$\boldsymbol{\mathrm{surface}}\:\boldsymbol{\mathrm{area}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{instant}}\:\boldsymbol{\mathrm{when}}\:\boldsymbol{\mathrm{the}} \\ $$$$\boldsymbol{\mathrm{radius}}\:\boldsymbol{\mathrm{is}}\:\mathrm{5}\boldsymbol{\mathrm{cm}}.\boldsymbol{\mathrm{What}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{radius}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}} \\ $$$$\boldsymbol{\mathrm{sphere}}\:\boldsymbol{\mathrm{when}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{surface}}\:\boldsymbol{\mathrm{area}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{increasing}} \\ $$$$\boldsymbol{\mathrm{at}}\:\mathrm{2}\boldsymbol{\mathrm{cm}}^{\mathrm{2}} /\boldsymbol{\mathrm{sec}}. \\ $$$$\boldsymbol{\mathrm{Any}}\:\boldsymbol{\mathrm{step}}\:\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{step}}\:\boldsymbol{\mathrm{solution}}\:\boldsymbol{\mathrm{pls}}. \\ $$

Answered by physicstutes last updated on 23/May/21

V = (4/3)πr^3 , (dV/dt) = 4πr^2 (dr/dt) 10 = 4π(5)^2 (dr/dt) ⇒ (dr/dt) = ((10)/(25(4π))) A = 4πr^2 ⇒ (dA/dt) = 8π(dr/dt)

$$\:{V}\:=\:\frac{\mathrm{4}}{\mathrm{3}}\pi{r}^{\mathrm{3}} \:,\:\:\:\frac{{dV}}{{dt}}\:=\:\mathrm{4}\pi{r}^{\mathrm{2}} \frac{{dr}}{{dt}} \\ $$$$\mathrm{10}\:=\:\mathrm{4}\pi\left(\mathrm{5}\right)^{\mathrm{2}} \frac{{dr}}{{dt}}\:\Rightarrow\:\frac{{dr}}{{dt}}\:=\:\frac{\mathrm{10}}{\mathrm{25}\left(\mathrm{4}\pi\right)} \\ $$$${A}\:=\:\mathrm{4}\pi{r}^{\mathrm{2}} \:\Rightarrow\:\frac{{dA}}{{dt}}\:=\:\mathrm{8}\pi\frac{{dr}}{{dt}} \\ $$

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